Question

Giải pt

Answer 1

Đề yêu cầu gì em?

 

Answer 2

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}=\dfrac{2x+1}{x\left(x+1\right)}\\ \Leftrightarrow x^2-1+x=2x+1\\ \Leftrightarrow x^2-1+x-2x-1=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x^2-2x\right)+\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)